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anonymous

5 years ago

@oldrin.bataku

anonymous

5 years ago

Indeed! Sorry, those are the actual extrema values. I gave you the arguments. Plug them into \(f(x,y,z)\):$$f(\dots)=\frac{9\sqrt3}{\sqrt{41}}+\frac{2\sqrt3}{\sqrt{41}}+\frac8{\sqrt{123}}=\frac{9(3)+2(3)+8}{\sqrt{123}}=\frac{41}{\sqrt{123}}=\frac{\sqrt{41}}{\sqrt3}=\sqrt{\frac{41}3}\approx3.7$$